Question Number 114176 by bemath last updated on 17/Sep/20

(1) 3x^2  ln (y) dx + (x^3 /y)dy = 0  (2) (e^(2x) +4)y ′= y   (3) dz = t(t^2 +1).e^(2z)  dt

Commented bybobhans last updated on 17/Sep/20

(2) (dy/y) = (dx/(e^(2x) +4)) ⇒∫ (dy/y) = ∫ (dx/(e^(2x) +4))  ⇒ ln (y) = ∫ (dx/(e^(2x) +4))  [ let e^x  = 2tan ψ →e^x  dx = 2sec ^2 ψ dψ ]  ⇔ ln (y) = ∫ ((2sec ^2 ψ )/(2tan ψ. 4sec ^2 ψ))dψ   ⇔ ln (y) = (1/4)∫ ((d(sin ψ))/(sin ψ))   ln (y) = (1/4)ln (sin ψ)+c    ln (y) = ln (C ((sin ψ))^(1/(4 ))  ) ⇒ y = C ((e^x /(4+e^(2x) )))^(1/(4 )) .

Answered by bobhans last updated on 17/Sep/20

Answered by Dwaipayan Shikari last updated on 17/Sep/20

∫(dz/e^(2z) )=∫t(t^2 +1)dt  −(1/2)e^(−2z) =(t^4 /4)+(t^2 /2)+C  e^(−2z) =−(t^4 /2)−t^2 +C_1   −2z=log(C_1 −(t^4 /2)−t^2 )  z=−(1/2)log(C_1 −(t^4 /2)−t^2 )  z=log((1/( (√(C_1 −(t^4 /2)−t^2 )))))

Answered by mathmax by abdo last updated on 18/Sep/20

2) (e^(2x)  +4)y^′  =y ⇒(y^′ /y) =(1/(e^(2x)  +4)) ⇒ln∣y∣ =∫ (dx/(e^(2x)  +4))  =_(e^x  =t)    ∫  (dt/(t(t^2  +4))) =(1/4)∫ ((1/t)−(t/(t^2  +4)))dt =(1/4)ln(t)−(1/8)ln(t^2  +4) +c  ⇒y(x) =k t^(1/4) .(t^2  +4)^(−(1/8))